leetcode26.Remove Duplicates from Sorted Array

问题描述：

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

思路：

   可以用一个下标 idx 代表新的数组的下标，初始值为0，遍历给定数组，当nums[i] != nums[idx]时，更新idx并赋新值。

注意：

   应当注意给出一个空数组的极限情况。

代码：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.size() < 2)
            return nums.size();
        int stride = 0;
        int idx = 0;
        for(int i = 0; i < nums.size(); i ++){
            if(nums[i] == nums[idx]){
                continue;
            }else{
                nums[++idx] = nums[i];
            }
        }
        return (idx+1);
    }
};